Question

A 5.00 gram bullet leaves the barrel of a horizontally aimed rifle located 1.00 [m]

above the ground with a speed of 250. [m/s]. (a) What is the force on the bullet

(assumed constant) as it travels down the 0.500 m barrel of the rifle? (b) How long

after leaving the barrel of the rifle is it before the bullet hits the ground? (c) How far

does the bullet travel before hitting the ground? Assume air resistance can be

neglected.

Answer 1

a)

From Kinematic equation

250² =0+2*a_x*0.5

a_x=62,500 m/s²

Force on the bullet is

F=ma_X=(5*10^-3)(62500)

F=312.5 N

b)

from Kinematic equation

Y=Y_o+V_oyt-(1/2)gt²

When bullet hits ground Y=0

0=1+0-(1/2)(9.8)t²

t=sqrt(2/9.8) =0.45 s

c)

Horizontal distance traveled

D=V_oxt =250*0.45

D=112.94 m =113 m (approx)